-0.000 282 005 906 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 906 8(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.000 282 005 906 8(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 282 005 906 8| = 0.000 282 005 906 8


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 282 005 906 8.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 282 005 906 8 × 2 = 0 + 0.000 564 011 813 6;
  • 2) 0.000 564 011 813 6 × 2 = 0 + 0.001 128 023 627 2;
  • 3) 0.001 128 023 627 2 × 2 = 0 + 0.002 256 047 254 4;
  • 4) 0.002 256 047 254 4 × 2 = 0 + 0.004 512 094 508 8;
  • 5) 0.004 512 094 508 8 × 2 = 0 + 0.009 024 189 017 6;
  • 6) 0.009 024 189 017 6 × 2 = 0 + 0.018 048 378 035 2;
  • 7) 0.018 048 378 035 2 × 2 = 0 + 0.036 096 756 070 4;
  • 8) 0.036 096 756 070 4 × 2 = 0 + 0.072 193 512 140 8;
  • 9) 0.072 193 512 140 8 × 2 = 0 + 0.144 387 024 281 6;
  • 10) 0.144 387 024 281 6 × 2 = 0 + 0.288 774 048 563 2;
  • 11) 0.288 774 048 563 2 × 2 = 0 + 0.577 548 097 126 4;
  • 12) 0.577 548 097 126 4 × 2 = 1 + 0.155 096 194 252 8;
  • 13) 0.155 096 194 252 8 × 2 = 0 + 0.310 192 388 505 6;
  • 14) 0.310 192 388 505 6 × 2 = 0 + 0.620 384 777 011 2;
  • 15) 0.620 384 777 011 2 × 2 = 1 + 0.240 769 554 022 4;
  • 16) 0.240 769 554 022 4 × 2 = 0 + 0.481 539 108 044 8;
  • 17) 0.481 539 108 044 8 × 2 = 0 + 0.963 078 216 089 6;
  • 18) 0.963 078 216 089 6 × 2 = 1 + 0.926 156 432 179 2;
  • 19) 0.926 156 432 179 2 × 2 = 1 + 0.852 312 864 358 4;
  • 20) 0.852 312 864 358 4 × 2 = 1 + 0.704 625 728 716 8;
  • 21) 0.704 625 728 716 8 × 2 = 1 + 0.409 251 457 433 6;
  • 22) 0.409 251 457 433 6 × 2 = 0 + 0.818 502 914 867 2;
  • 23) 0.818 502 914 867 2 × 2 = 1 + 0.637 005 829 734 4;
  • 24) 0.637 005 829 734 4 × 2 = 1 + 0.274 011 659 468 8;
  • 25) 0.274 011 659 468 8 × 2 = 0 + 0.548 023 318 937 6;
  • 26) 0.548 023 318 937 6 × 2 = 1 + 0.096 046 637 875 2;
  • 27) 0.096 046 637 875 2 × 2 = 0 + 0.192 093 275 750 4;
  • 28) 0.192 093 275 750 4 × 2 = 0 + 0.384 186 551 500 8;
  • 29) 0.384 186 551 500 8 × 2 = 0 + 0.768 373 103 001 6;
  • 30) 0.768 373 103 001 6 × 2 = 1 + 0.536 746 206 003 2;
  • 31) 0.536 746 206 003 2 × 2 = 1 + 0.073 492 412 006 4;
  • 32) 0.073 492 412 006 4 × 2 = 0 + 0.146 984 824 012 8;
  • 33) 0.146 984 824 012 8 × 2 = 0 + 0.293 969 648 025 6;
  • 34) 0.293 969 648 025 6 × 2 = 0 + 0.587 939 296 051 2;
  • 35) 0.587 939 296 051 2 × 2 = 1 + 0.175 878 592 102 4;
  • 36) 0.175 878 592 102 4 × 2 = 0 + 0.351 757 184 204 8;
  • 37) 0.351 757 184 204 8 × 2 = 0 + 0.703 514 368 409 6;
  • 38) 0.703 514 368 409 6 × 2 = 1 + 0.407 028 736 819 2;
  • 39) 0.407 028 736 819 2 × 2 = 0 + 0.814 057 473 638 4;
  • 40) 0.814 057 473 638 4 × 2 = 1 + 0.628 114 947 276 8;
  • 41) 0.628 114 947 276 8 × 2 = 1 + 0.256 229 894 553 6;
  • 42) 0.256 229 894 553 6 × 2 = 0 + 0.512 459 789 107 2;
  • 43) 0.512 459 789 107 2 × 2 = 1 + 0.024 919 578 214 4;
  • 44) 0.024 919 578 214 4 × 2 = 0 + 0.049 839 156 428 8;
  • 45) 0.049 839 156 428 8 × 2 = 0 + 0.099 678 312 857 6;
  • 46) 0.099 678 312 857 6 × 2 = 0 + 0.199 356 625 715 2;
  • 47) 0.199 356 625 715 2 × 2 = 0 + 0.398 713 251 430 4;
  • 48) 0.398 713 251 430 4 × 2 = 0 + 0.797 426 502 860 8;
  • 49) 0.797 426 502 860 8 × 2 = 1 + 0.594 853 005 721 6;
  • 50) 0.594 853 005 721 6 × 2 = 1 + 0.189 706 011 443 2;
  • 51) 0.189 706 011 443 2 × 2 = 0 + 0.379 412 022 886 4;
  • 52) 0.379 412 022 886 4 × 2 = 0 + 0.758 824 045 772 8;
  • 53) 0.758 824 045 772 8 × 2 = 1 + 0.517 648 091 545 6;
  • 54) 0.517 648 091 545 6 × 2 = 1 + 0.035 296 183 091 2;
  • 55) 0.035 296 183 091 2 × 2 = 0 + 0.070 592 366 182 4;
  • 56) 0.070 592 366 182 4 × 2 = 0 + 0.141 184 732 364 8;
  • 57) 0.141 184 732 364 8 × 2 = 0 + 0.282 369 464 729 6;
  • 58) 0.282 369 464 729 6 × 2 = 0 + 0.564 738 929 459 2;
  • 59) 0.564 738 929 459 2 × 2 = 1 + 0.129 477 858 918 4;
  • 60) 0.129 477 858 918 4 × 2 = 0 + 0.258 955 717 836 8;
  • 61) 0.258 955 717 836 8 × 2 = 0 + 0.517 911 435 673 6;
  • 62) 0.517 911 435 673 6 × 2 = 1 + 0.035 822 871 347 2;
  • 63) 0.035 822 871 347 2 × 2 = 0 + 0.071 645 742 694 4;
  • 64) 0.071 645 742 694 4 × 2 = 0 + 0.143 291 485 388 8;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 282 005 906 8(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0101 1010 0000 1100 1100 0010 0100(2)

6. Positive number before normalization:

0.000 282 005 906 8(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0101 1010 0000 1100 1100 0010 0100(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 282 005 906 8(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0101 1010 0000 1100 1100 0010 0100(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0101 1010 0000 1100 1100 0010 0100(2) × 20 =

1.0010 0111 1011 0100 0110 0010 0101 1010 0000 1100 1100 0010 0100(2) × 2-12


8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -12

Mantissa (not normalized):
1.0010 0111 1011 0100 0110 0010 0101 1010 0000 1100 1100 0010 0100


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(10)


10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1011(10) =

011 1111 0011(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 0010 0111 1011 0100 0110 0010 0101 1010 0000 1100 1100 0010 0100 =

0010 0111 1011 0100 0110 0010 0101 1010 0000 1100 1100 0010 0100


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 0011

Mantissa (52 bits) =
0010 0111 1011 0100 0110 0010 0101 1010 0000 1100 1100 0010 0100


Decimal number -0.000 282 005 906 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 0101 1010 0000 1100 1100 0010 0100

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100